- 91.
- A square and a regular pentagon are inscribed in a circle. The nine vertices are all distinct and divide the circumference into nine arcs. Prove that at least one of them does not exceed 1/40 of the circumference of the circle.

Suppose a regular pentagon is inscribed in a circle. Since any
pair of adjacent vertices terminate a closed arc whose length
is 1/5 of the circumference, no two vertices of the pentagon
can belong to the same longer arc. Since there are only four
longer arcs, at least one vertex of the pentagon must lie
inside a shorter arc and so be no more distant that
$\frac{1}{2}\times \frac{1}{20}=\frac{1}{40}$ from
a vertex of the square.

- 92.
- Consider the sequence $200125$, $2000125$, $20000125$, $\dots $, $200\dots 00125$, $\dots $ (in which the $n$th number has $n+1$ digits equal to zero). Prove that none of these numbers is the square, cube or fifth power of an integer.

Observe that
${10}^{n+1}={10}^{n-2}\times 8\times 125$, so that

If such a number is a $k$th power, it must be divisible by ${5}^{k}$. Since $16\times {10}^{n-2}+1$ is not a multiple of 5, no number in the sequence can be a $k$th power for $k\ge 4$.

Let
${u}_{n}=16\times {10}^{n-2}+1$, for
$n\ge 5$. Suppose
that
$(10{x}_{n}+1){}^{3}={u}_{n}$. Then

so that ${x}_{n}={10}^{n-3}{y}_{n}$ for some ${y}_{n}$ and we find that

so that ${y}_{n}=1$ or 2. It is straightfoward to check that neither works, so that ${u}_{n}$ can never be a cube. Hence no number in the given sequence can be a cube.

- 93.
- For any natural number $n$, prove the following inequalities:

whence $s=2-(n+2){2}^{-n}$. Clearly, $s<2$.

On the other hand,

When $n=1$, ${2}^{n+1}-(5n-2)=1>0$; when $n=2$, ${2}^{n+1}-(5n-2)=0$, and when $n=3$, ${2}^{n+1}-(5n-2)=3>0$. Suppose, as an induction hypothesis, that ${2}^{k+1}>5k-2$ for some $k\ge 3$. Then

so that, for each positive integer $n$, ${2}^{n+1}\ge 5n-2$, with equality if and only if $n=2$. The desired result follows.

- 94.
- $\mathrm{ABC}$ is a right triangle with arms $a$ and $b$ and hypotenuse $c=\Vert \mathrm{AB}\Vert $; the area of the triangle is $s$ square units and its perimeter is $2p$ units. The numbers $a$, $b$ and $c$ are positive integers. Prove that $s$ and $p$ are also positive integers and that $s$ is a multiple of $p$.

- 95.
- The triangle $\mathrm{ABC}$ is isosceles is isosceles with equal sides $\mathrm{AC}$ and $\mathrm{BC}$. Two of its angles measure ${40}^{\u02c6}$. The interior point $M$ is such that $\angle \mathrm{MAB}={10}^{\u02c6}$ and $\angle \mathrm{MBA}={20}^{\u02c6}$. Determine the measure of $\angle \mathrm{CMB}$.

By the Law of Sines,

and

where $\alpha =\angle \mathrm{CMB}$. Hence

whence

Since $\alpha +{40}^{\u02c6}=(\alpha -{50}^{\u02c6})+{90}^{\u02c6}$, $\mathrm{sin}(\alpha +{40}^{\u02c6})=\mathrm{cos}(\alpha -{50}^{\u02c6})$. Therefore, we find that $2\mathrm{cos}(\alpha -{50}^{\u02c6})\mathrm{sin}{10}^{\u02c6}=\mathrm{cos}(\alpha -{50}^{\u02c6})$. Since $\mathrm{sin}{10}^{\u02c6}\ne \frac{1}{2}$, we have that $\mathrm{cos}(\alpha -{50}^{\u02c6})=0$ and $\alpha ={140}^{\u02c6}$.

- 96.
- Find all prime numbers $p$ for which all three of the numbers ${p}^{2}-2$, $2{p}^{2}-1$ and $3{p}^{2}+4$ are also prime.